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Breaking Brackets Demo 3(2c+5). Single Bracket 0 123 456 789 Ans5 C. ÷ x 0 + On ² - Ans = √ (-) 1 ( - ) x8 = - 2 3 4 5 6 7 8 9 10 Show Working Clear r.

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Presentation on theme: "Breaking Brackets Demo 3(2c+5). Single Bracket 0 123 456 789 Ans5 C. ÷ x 0 + On ² - Ans = √ (-) 1 ( - ) x8 = - 2 3 4 5 6 7 8 9 10 Show Working Clear r."— Presentation transcript:

1 Breaking Brackets Demo 3(2c+5)

2 Single Bracket 0 123 456 789 Ans5 C. ÷ x 0 + On ² - Ans = √ (-) 1 ( - ) x8 = - 2 3 4 5 6 7 8 9 10 Show Working Clear r

3 How to “Break” Brackets 2 Simply multiply everything inside the bracket by what is outside the bracket x( 4 +) = 2 x x + 4 = 8 2x2x2x2x = 8 The + between the x and 4 tells you to add the answers + 2x2x2x2x Two “answers one for each term inside the brackets

4 How to “Break” Brackets 5 Simply multiply everything inside the bracket by what is outside the bracket x( 3 -) = 5 x x - 3 = 15 5x5x5x5x = 15 The - between the x and 3 tells you to subtract the answers - 5x5x5x5x Two “answers one for each term inside the brackets

5 How to “Break” Brackets 4444 Simply multiply everything inside the bracket by what is outside the bracket rrrr( 7777- ) = 0 123 456 789 C. ÷ x 0 + On ² - Ans = √ (-) rx - x = - New Example Start

6 More Examples 2 ( a + 3 ) = Examples Multiply Out 2a+ 6 5 ( b + 3 ) = 5b+ 15 8 ( c + 7 ) = 8c+ 56 3 ( d - 2 ) = 3d- 6 9 ( e - 5 ) = 9e- 45 (a) 7 ( f + 4 ) = 7f + 28 (b) 6 ( g - 5 ) = 6g - 30 (c) 5 ( h - 8 ) = 5h - 40 (d) 10 ( k + 7 ) = 10k + 70 (e) 9 ( m - 8 ) = 9m - 72 (f) 4 ( n + 11 ) = 4n + 44

7 3(a+5) type Examples a) b) c) d) e) f) g) 4(a-9) = 4a-36 3(a+5) = 7(a+3) = 5(a+4) = 6(a-7) = 9(a-2) = 8(a-6) = 3a+15 7a+21 5a+20 6a-42 9a-18 8a-48 0 123 456 789 C. ÷ x 0 + On ² - Ans = √ (-) a New Examples Show Answers

8 3(2c+5) 3 Simply multiply everything inside the bracket by what is outside the bracket 2c ( 5 ) = 3 x 2c 3 x + 5 = 15 6 x c = 15 The + between the 2c and 5 tells you there is a + between the answers + 6c 2 x c x 3 Two answers, one for each term x 3

9 2(5z-7) 2 Simply multiply everything inside the bracket by what is outside the bracket 5z ( 7 ) = 2 x 5z 2 x - 7 = 14 10 x z = 14 The - between the 5z and 7 tells you there is a - between the answers - 10z 5 x z x 2 Two answers, one for each term x 2

10 1 st term in brackets 3b, 2a, 7g etc 7777 Simply multiply everything inside the bracket by what is outside the bracket 2q ( 2222+ ) = 0 123 456 789 14 C. ÷ x 0 + On ² - Ans = √ (-) qx + x = + New Example Start

11 Examples like 5(3p – 2 ) 2 ( 2a + 3 ) = Examples Multiply Out 4a+ 6 4 ( 3b + 5 ) = 12b + 20 6 ( 2c + 3 ) = 12c+ 18 5 (5e - 4 ) = 25e- 20 9 ( 7e - 3 ) = 63e- 27 (a) 7 ( 2f + 5 ) = 14f + 35 (b) 6 ( 2g - 5 ) = 12g - 30 (c) 5 ( 7h - 9 ) = 35h - 45 (d) 10 ( 2k + 7 ) = 20k + 70 (e) 9 ( 3m - 9 ) = 27m - 81 (f) 4 ( 4n + 10 ) = 16n + 40

12 3(?a+5) type Examples a) b) c) d) e) f) g) 4(5a+4) = 20a+16 8(7a+6) = 2(9a-7) = 6(8a+9) = 9(4a+3) = 3(6a+5) = 7(2a+5) = 56a+48 18a-14 48a+54 36a+27 18a+15 14a+35 0 123 456 789 C. ÷ x 0 + On ² - Ans = √ (-) a New Examples Show Answers

13 Removing Brackets : 3(2a+b-5) 3 Simply multiply everything inside the bracket by what is outside the bracket 2a ( b ) = 3 x 2a 3 x + b = 3b 6 x a = 3b Copy the signs between each term + 6a 2 x a x 3 Three answers, one for each term x 3 5 - - b x 3 3 x 5 =15

14 Removing Brackets : 3(s + 4)+2s 3 Simply multiply everything inside the bracket by what is outside the bracket s ( 4 ) = + The + between the p and 3 tells you there is a + between the answers + x 3 Two answers, one for each term x 3 + 2s All Maths “sums” involve BODMAS. ( B ) first 3 x s = 3s 3 x 4 = 12

15 Removing Brackets : 3(s + 4)+2s 3 Simply multiply everything inside the bracket by what is outside the bracket s ( 4 ) = + + + 2s 3s 12 + 2s Three terms now The brackets equal 5p + 15 The middle is a number, the others letters. ….. can add the two letter terms 3s + 2s = 5s = + 12 The expression has been simplified to 5s + 12 A letter term plus a number term …… hide the number 5s

16 More Examples 5 ( d + 3 ) +d = Examples Multiply Out 5d + 15 + d = 6d + 15 7 ( e + 4 ) – 3e = 7e + 28 - 3e = 4e + 28 8 (f + 3 ) – 5f = 8f + 24 - 5f = 3f + 24 (a) 7(g + 3 ) + 2g = 9g + 21 (b) 4(h + 2 ) - h = 3h + 8 (c) 2(j + 7) + 3j = 5j + 14 (d) 9(k + 3 ) – 2k = 7k + 27 (e) 6(m + 9 ) – 4m = 2m + 54 (f) 8(n – 3 ) + 2n = 10n - 24

17 3(a+5) +5a type Examples a) b) c) d) e) f) g) 4(a-7) + 4a = 8a-28 2(a-4) + 5a = 5(a+8) + 7a = 9(a+5) + 6a = 8(a+3) + 8a = 6(a+6) + 2a = 3(a-2) + 9a = 7a-8 12a+40 15a+45 16a+24 8a+36 12a-6 0 123 456 789 C. ÷ x 0 + On ² - Ans = √ (-) a New Examples Show Answers 4a-28 + 4a = 4a-28 + 4a = 2a-8 + 5a = 2a-8 + 5a = 5a+40 + 7a = 5a+40 + 7a = 9a+45 + 6a = 9a+45 + 6a = 8a+24 + 8a = 8a+24 + 8a = 6a+36 + 2a = 6a+36 + 2a = 3a-6 + 9a = 3a-6 + 9a =

18 Removing Brackets : 5( p + 3 ) + 7 5 Simply multiply everything inside the bracket by what is outside the bracket p ( 3 ) = + The + between the p and 3 tells you there is a + between the answers + x 5 Two answers, one for each term x 5 + 7 All Maths “sums” involve BODMAS. ( B ) first 5 x p = 5p 5 x 3 = 15

19 Removing Brackets : 5( p + 3 ) + 7 5 Simply multiply everything inside the bracket by what is outside the bracket p ( 3 ) + + + 7 5p 15 + 7 Three terms now The brackets equal 5p + 15 The 1 st is a letter term the others numbers ….. can add the two numbers 15 + 7 = 22 5p = + 22 The expression has been simplified to 5p + 22 A letter term plus a number term

20 Some more examples 3 ( a + 4 ) +5 = Examples Multiply Out 3a + 12 + 5 = 3a + 17 5 ( b + 3 ) - 6 = 5b + 15 - 6 = 5b + 9 6 (c + 9 ) – 10 = 6c + 54 - 10 = 6c + 48 (a) 4(e + 5 ) + 3 = 4e + 23 (b) 2(f + 7 ) - 2 = 2f + 12 (c) 5(g + 3) + 10 = 5g + 25 (d) 3(h + 8 ) - 7 = 3h + 17 (e) 6(k + 2 ) - 11 = 6k + 1 (f) 8(m – 3 ) + 35 = 8m - 11

21 3(a+5) +7 type Examples a) b) c) d) e) f) g) 2(a-4) + 5 = 2a-3 6(a-7) + 4 = 4(a-3) + 2 = 3(a+5) + 7 = 5(a-2) + 6 = 8(a-6) + 3 = 9(a+8) + 9 = 6a-38 0 123 456 789 C. ÷ x 0 + On ² - Ans = √ (-) a New Examples Show Answers 2a-8 + 5 = 2a-8 + 5 = 6a-42 + 4 = 6a-42 + 4 =

22 Removing Brackets : 4( 2a + 5 ) + 6 4 Simply multiply everything inside the bracket by what is outside the bracket 2a ( 7 ) 4 x 2a 4 x + 7 = 28 8 x a = 28 The + between the 2a and 7 tells you there is a + between the answers + 8a 2 x a x 4 Two answers, one for each term x 4 + 6 All Maths “sums” involve BODMAS. ( B ) first

23 Removing Brackets : 4( 2a + 5 ) + 6 4 Simply multiply everything inside the bracket by what is outside the bracket 2a ( 7 ) = + + + 6 8a 28 + 6 Three terms now The brackets equal 8a + 28 The 1 st is a letter term the others numbers ….. can add the two numbers 28 + 6 = 34 8a = + 34 The expression has been simplified to 8a + 34 A letter term plus a number term

24 More Examples 3 ( 2a + 3 ) +4 = Examples Multiply Out 6a + 9 + 4 = 6a + 13 4 ( 3b + 5 ) - 3 = 12b + 20 - 3 = 12b + 17 6 ( 2c + 3 ) – 10 = 12c + 18 - 10 = 12c + 8 (a) 4(3e + 2 ) +9 = 12e + 17 (b) 2(7f + 3 ) -2 = 14f +4 (c) 5(2g + 5) + 6 = 10g + 31 (d) 3(5h + 8 ) - 7 = 15h + 17 (e) 6(3k + 9 ) -11 = 18k +43 (f) 8(2m – 3 ) + 10 = 16m - 14

25 Removing Brackets : 5( 3b - 7 ) + 4b 5 Simply multiply everything inside the bracket by what is outside the bracket 3b ( 7 ) = 5 x 3b 5 x - 7 = 35 15 x b = 35 The - between the 3b and 7 tells you there is a - between the answers - 15b 3 x b x 5 Two answers, one for each term x 5 + 4b All Maths “sums” involve BODMAS. ( B ) first

26 Removing Brackets : 5( 3b - 7 ) + 4b 5 Simply multiply everything inside the bracket by what is outside the bracket 3b ( 7 ) = - - + 4b 15b 35 + 4b Three terms now The brackets equal 15b -35 The 1 st and last terms are letter terms the middle a number Hide the middle term and add the two letter terms 15b +4b = 19b = - 35 The expression has been simplified to 19b - 35 A letter term plus a number term 19b

27 Removing Brackets : 3( 2a + 5b ) – 2a 3 Simply multiply everything inside the bracket by what is outside the bracket 2a ( 5b ) = 3 x 2a + 5b = 6 x a The + between the 2a and 5b tells you there is a + between the answers + 6a 2 x a x 3 Two answers, one for each term x 3 - 2a All Maths “sums” involve BODMAS. ( B ) first 3 x = 15 x b 15b 5 x b

28 Removing Brackets : 3( 2a + 5b ) – 2a 3 Simply multiply everything inside the bracket by what is outside the bracket 2a ( 5b ) + + - 2a 6a 15b - 2a Three terms now The brackets equal 6a +15b The 1 st and last terms are letter terms in a the middle a letter terms in b Hide the middle term and add the two terms in a 9a – 2a = 4g = + 45b The expression has been simplified to 4a +15b Two terms in different letters 4a

29 More Examples 4 ( 3a + 5 ) + a = Examples Multiply Out 12a + 20 + a = 13a + 20 3 ( 2g + 5 ) – 2g = 6g + 15 - 2g = 4g + 15 5 ( 3s + 2t ) – 6s = 15s +10t - 6s = 9s + 10t (a) 5(3e + 2 ) +2e = 17e + 10 (b) 3(4f + 5 ) – 7f = 5f + 15 (c) 6(3g + 2) + 4g = 22g +12 (d) 2(5h + 3g ) – 6h = 4h +6g (e) 4(5s + 9t ) – 3t = 20s +33t (f) 8(2m – 5a ) – 3m = 13m -40a

30 Removing Brackets : 7a + 3(2a – 5) 3 Simply multiply everything inside the bracket by what is outside the bracket 2a ( 5 ) = 3 x 2a 3 x - 5 = 15 6 x a = 15 The - between the 2a and 5 tells you there is a - between the answers - 6a 2 x a x 3 Two answers, one for each term x 3 7a + All Maths “sums” involve BODMAS. ( B ) first

31 Removing Brackets : 7a + 3(2a – 5) 5 Simply multiply everything inside the bracket by what is outside the bracket 3b ( 7 ) = - - 4b 15b 35 4b Three terms now The brackets equal 15b -35 The 1 st and last terms are letter terms the middle a number Hide the middle term and add the two letter terms 15b +4b = 19b = - 35 The expression has been simplified to 19b - 35 A letter term plus a number term 19b + + 4b

32 More Examples 2a + 4 ( 5a + 1 ) = Examples Multiply Out 20a + 4 2a + 13a + 20 5g +2 ( 3g - 4 ) = 6g - 8 5g + 11g - 8 4s +3 ( 6s + 2t ) = 18s +6t 4s + 22s + 6t (a) 4e+3(2e + 7 ) = 10e + 21 (b) 3g + 5(2g + 1) = 13g + 5 (c) 5b + 2(9b – 4) = 23b - 8 (d) 9t +4(3t – 5) = 21t - 20 (e) 6r + 2(3r + 8) = 12r + 16 (f) 10f + 3(9 – 2f) = 4f + 27 = = =


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